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2 years ago

If vertex A is at (-1, 2) and vertex B is at (1, 5), then vertex A' is at and vertex B' is at .

Mathematics

1 answer:

Serggg [28]2 years ago

8 0

Answer:

Vertical translation is along the y axis while horizontal translation is along the x-axis.

Vertex A' is at (-1 - 3, 2 + 6) = (-4, 8)

Vertex B' is at (1 - 3, 5 + 6) = (-2, 11)

Brainliest Pls

Step-by-step explanation:

You might be interested in

Which calculation will ALWAYS give a result greater than 1? A.5 × a number less than 1 B.4/9+ a fraction less than 12 C. 1 3/4 -

Bas_tet [7]

Answer:

C. 1 3/4 - (a fraction less than 3/4)

Step-by-step explanation:

Your number sense should be able to help you with this one.

A. 5 × 1/10 = 1/2, not a number greater than 1

B. 4/9 + 1/3 = 7/9, not a number greater than 1

C. 1 3/4 -1/2 = 1 1/4, a number greater than 1 (see below for more explanation)

D. 7/8 × 1/2 = 7/16, not a number greater than 1.

More explanation

Let x represent a number less than 3/4. Then we want to make sure that ...

y = 1 3/4 - x

will be greater than 1.

Solving for x, we get ...

x = 1 3/4 - y

Applying the requirement that x < 3/4, we have ...

x < 3/4

(1 3/4 -y) < 3/4

1 3/4 < y + 3/4 . . . . . . add y

1 < y . . . . . . . . . . . . . . .subtract 3/4

We see that the condition on x makes sure that y is always greater than 1.

3 0

3 years ago

Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :

$\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}$

$k \approx \pm 0.028$

The positive value of $k$ will result in exactly one real root is approximately 0.028.

7 0

2 years ago

There is 1/4 ounce of yeast in every 2 1/4 teaspoon of yeast. A recipe calls for 2 teaspoons of yeast. How many ounces of yeast

MrRissso [65]

1/4 ounce of yeast ... 2 1/4 teaspoons of yeast
x ounce of yeast = ? ... 2 teaspoons of yeast

If you would like to know how many ounces of yeast need to be in the recipe, you can calculate this using the following steps:

1/4 * 2 = x * 2 1/4
1/2 = x * 9/4 /*4/9
x = 1/2 * 4/9
x = 2/9

Result: 2/9 ounce of yeast needs to be in this recipe.

7 0

3 years ago

5^(2x-1)+5^(x+1)=250 how do you solve? thank you

zepelin [54]

Answer:

x = 2

Step-by-step explanation:

Exponential Equations

Solve:

$5^{2x-1}+5^{x+1}=250$